1. Chapter 5 Class 10 Arithmetic Progressions
2. Serial order wise
3. Ex 5.3

Transcript

Ex 5.3 ,4 How many terms of the AP. 9, 17, 25 โฆ must be taken to give a sum of 636? AP is 9,17,25, โฆ.. We know that Sum = ๐/2 (2a + (n โ 1) d) Here, a = 9 d = 17 โ 9 = 8 & Sum = Sn = 636 We need to find n Putting values in equation Sum = ๐/2 (2a + (n โ 1) d) 636 = ๐/(2 ) (2 ร 9 + (n โ 1) ร 8) 636 ร 2 = n (18 + (n โ 1) ร 8) 1272 = n (18 + 8n โ 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 โ 1272 10n + 8n2 โ 1252 = 0 2(5n + 4n2 โ 636)= 0 5n + 4n2 โ 636 = 0 4n2 + 5n โ 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = โ 636 We know that, D = b2 โ 4ac D = (5)2 โ 4 ร 4 ร (โ636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (โ ๐ ยฑ โ๐ท)/2๐ Putting values n = (โ 5 ยฑ โ10201)/(2 ร 4) n = (โ 5 ยฑ 101)/8 Solving So, n = 12 & n = (โ53)/4 Since n cannot be negative, n = 12

Ex 5.3